In the circuit shown,$C_1 = 6\,\mu F, C_2 = 3 \,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge (in $\mu C$) finally on $C_2$?

Two identical capacitors are joined in parallel,charged to a potential $V$ and then separated and then connected in series,i.e.,the positive plate of one is connected to the negative plate of the other. What happens to the system?

$A$ $60\; pF$ capacitor is fully charged by a $20\; V$ supply. It is then disconnected from the supply and connected to another uncharged $60\; pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $nJ$):

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

$A$ $0.2 \, F$ capacitor is charged to $600 \, V$ by a battery. After removing the battery,it is connected to another uncharged parallel plate capacitor of $1 \, F$. The potential decreases to $V$ equal to:

$A$ capacitor of capacitance $0.2 \, \mu F$ is charged to a potential of $600 \, V$. After removing the battery,it is connected in parallel to an uncharged capacitor of $1.0 \, \mu F$. The final potential across the capacitors will be.........$V$.